Question

For a prism kept in air, it is found that for an angle of incidence ${60}^{\circ }$, the angle of refraction ${}^{\prime }{A}^{\prime }$, angle of deviation ${}^{\prime }{\delta }^{\prime }$ and angle of emergence ${}^{\prime }{e}^{\prime }$ become equal. Then, the refractive index of the prism is

• A. $1.73$
• B. $1.15$
• C. $1.5$
• D. $1.33$

Answer 1

The correct option is A $1.73$

Given $i={60}^{\circ }A=\delta =e$
$\delta =i+e-A\Rightarrow \delta =1(\because e=A)$
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
Here, angle of deviation is minimum $(\because i=e)$
$\mu =\frac{\sin \left(\frac{{60}^{\circ }+{60}^{\circ }}{2}\right)}{\sin \left({60}^{\circ }/2\right)}=1.73$.